Density of `2.05 M` solution of acetic acid in water is `1.02g//mL`. The molality of same solution is:

Options

A. `1.14 mol kg^(-1)`
B. `3.28 mol kg^(-1)`
C. `2.28 mol kg^(-1)`
D. `0.44 mol kg^(-1)`

Answer

C

Explanation

`"Molality"=("Moles of"CH_(3)COOH)/("wt.of solvent in" kg)=(2.05xx1000)/(897)`
`2.285 mol kh^(-1)`
`[{:("wt.of solvent=","wt.of solution"-),(,"wt.of solute"),(=[1000xx1.02]-,[2.05xx60]),(=897g,):}]`